### 28th Putnam 1967

Problem B2

A, B ∈ [0, 1] and we have ax2 + bxy + cy2 ≡ (A x + (1 - A )y)2, (A x + (1 - A)y )(B x + (1 - B)y) ≡ dx2 + exy + fy2. Show that at least one of a, b, c ≥ 4/9 and at least one of d, e, f ≥ 4/9.

Solution

For the first part, a = A2, b = 2 A (1 - A), c = (1 - A)2. If a < 4/9, then A < 2/3. If c < 4/9 then A > 1/3. But b > 4/9 for 1/3 < A < 2/3.

For the second part, d = A B, f = (1 - A )(1 - B ), e = A + B - 2 AB.

Measure A along the x-axis and B along the y-axis. Consider the regions of the square for which each of d, e, f are ≥ 4/9. d ≥ 4/9 for points above the hyperbola y = 4/(9x) which passes through the points (4/9, 1), (2/3, 2/3), (1, 4/9). Similarly f ≥ 4/9 for points below the hyperbola y = 1 - 4/(9 - 9x), which passes through the points (0, 5/9), (1/3, 1/3), (5/9, 0). Finally, e > 4/9 for points lying between the two branches of the hyperbola y = (4 - x)/(9 - 18x). The bottom branch passes through (0, 4/9), (1/3, 1/3), (4/9, 0) and the top branch passes through (5/9, 1), (2/3, 2/3), (1, 5/9). Thus the bottom branch is entirely below the d = 4/9 hyperbola, except at the point of intersection (1/3, 1/3), and the top branch is entirely above the f = 4/9 hyperbola, except at the point of intersection (2/3, 2/3). This is easily checked by solving d = e = 4/9 and d = f = 4/9. Thus the three areas d ≥ 4/9, e ≥ 4/9, f ≥ 4/9 cover the unit square.